2016-03-07 108 views
0

我希望看到发送请求数据包之前,我发送它作为req中有一个错误和api是一般性描述500 错误,所以我不能告诉y请求失败。我知道xml的格式不正确,因为它在Chrome上运行在postman上。查看guzzle发帖请求

$client = new GuzzleHttp\Client([ 
      'base_uri' => 'https://elstestserver.endicia.com', 
     ]); 
$xml = 'changePassPhraseRequestXML=<ChangePassPhraseRequest> <RequesterID>lxxx</RequesterID><RequestID>1263055835</RequestID><CertifiedIntermediary><AccountID>lxxx</AccountID><PassPhrase>dfdsfsd</PassPhrase></CertifiedIntermediary><NewPassPhrase>fdfdsfdsfs</NewPassPhrase></ChangePassPhraseRequest>'; 

    $data = array("ChangePassPhraseXML" => $xml); 

    $response = $client->post("/LabelService/EwsLabelService.asmx/ChangePassPhraseXML", [ 
'form_params' => $data 
]); 

申请工作在邮递员铬的继承人的XML的工作示例

changePassPhraseRequestXML=<ChangePassPhraseRequest><RequesterID>lxxx</RequesterID><RequestID>1263055835</RequestID><CertifiedIntermediary><AccountID>lxxx</AccountID><PassPhrase>dfdsfsd</PassPhrase></CertifiedIntermediary><NewPassPhrase>fdsfdsfds</NewPassPhrase></ChangePassPhraseRequest> 
+0

你会寻找调试请求选项。如果在客户端实例化中将其设置为true,则客户端发出的所有请求都会将请求输出到php:// stdout。供参考:http://guzzle.readthedocs.org/en/latest/request-options.html#debug –

回答